A glass tube of 100 cm length is filled with water. The water can be drained out slowly at the bottom of the tube. If a vibrating tuning fork of frequency 500 Hz is brought at the upper end of the tube and the velocity of sound in air is 330 m/s, then the total number of resonances obtained will be4123
Question
A glass tube of 100 cm length is filled with water. The water can be drained out slowly at the bottom of the tube. If a vibrating tuning fork of frequency 500 Hz is brought at the upper end of the tube and the velocity of sound in air is 330 m/s, then the total number of resonances obtained will be4123
Solution
The problem is about the resonance in a tube open at one end and closed at the other end. The formula for the fundamental frequency (or first harmonic) in such a tube is given by:
v = 4fL
where: v is the speed of sound, f is the frequency, and L is the length of the tube.
We can rearrange this formula to solve for L:
L = v / (4f)
Given that v = 330 m/s and f = 500 Hz, we can substitute these values into the formula:
L = 330 / (4 * 500) = 0.165 m = 16.5 cm
This is the length of the tube required for the first resonance. The length of the tube for the nth resonance is given by:
L_n = (2n - 1) * L / 2
The total number of resonances is the largest integer n such that L_n ≤ 100 cm. We can solve this inequality for n:
(2n - 1) * 16.5 ≤ 100 2n - 1 ≤ 100 / 16.5 2n ≤ 100 / 16.5 + 1 n ≤ (100 / 16.5 + 1) / 2
Solving this inequality gives n ≤ 3.03. Therefore, the total number of resonances is 3.
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